# Need help re-arranging an equation!

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• Jun 19th 2012, 05:49 AM
rowe
Re: Need help re-arranging an equation!
Remember the rule $\frac{a^2}{b^2} = {\left(\frac{a}{b}\right)}^2$? well, if you square both sides of the equation, you'll remove the square root.

Then, you can rearrange to remove the fraction, and solving for $z$ should be easier.
• Jun 19th 2012, 06:08 AM
Rhys101
Re: Need help re-arranging an equation!
Rowe! Your back! I thought you might have gone to bed. :)Not sure what time zone you are in. Anyway! do I end up with: $x^2=(MR/m)^2*((r^2)^2/((1+r^2)^(2+2zr)^2))$
or am I completely off track again?
• Jun 19th 2012, 08:31 AM
rowe
Re: Need help re-arranging an equation!
Almost, if my calculations are correct (this is quite a tedious one!) you should have $x^2 = \left(\frac{MR}{m}\right)^2 \cdot \frac{r^4}{{(1-r^2)}^2 + (2zr)^2}$. You want to try and isolate $z$, start by getting rid of that fraction by multiplying both sides by ${(1-r^2)}^2 + (2zr)^2$. Once that's "on the flat", simplify / expand where necessary so you can subtract the $(2zr)^2$ term on its own side of the equation and work from there.
• Jun 19th 2012, 08:55 AM
Wilmer
Re: Need help re-arranging an equation!
Quote:

Originally Posted by rowe
$x = \frac{MR}{m}\cdot\frac{r^2}{\sqrt{{(1-r^2)}^2+{(2zr)}^2}}$

Looks like this is the final/accepted version (Clapping)

Let k = SQRT[(1 - r^2)^2 + (2zr)^2] ; then:

x = MRr^2 / (mk)
k = MRr^2 / (mx) ; square both sides:

k^2 = M^2 R^2 r^4 / (m^2 x^2) ; substitute back in:

(1 - r^2)^2 + (2zr)^2 = M^2 R^2 r^4 / (m^2 x^2) ; rearrange:

4 z^2 r^2 = M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2

z^2 = [M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2] / (4r^2) ; sooooooooooo:

z = SQRT{[M^2 R^2 r^4 / (m^2 x^2) - (1 - r^2)^2] / (4r^2)}

Can be slightly simplified to:
z = SQRT[M^2 R^2 r^4 - (m^2 x^2)(1 - r^2)^2] / (2rmx)
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